Problem: If $x \bigtriangleup y = 4x-6y$ and $x \circledcirc y = 3x+y$, find $(1 \bigtriangleup 0) \circledcirc 0$.
First, find $1 \bigtriangleup 0$ $ 1 \bigtriangleup 0 = (4)(1)-(6)(0)$ $ \hphantom{1 \bigtriangleup 0} = 4$ Now, find $4 \circledcirc 0$ $ 4 \circledcirc 0 = (3)(4)$ $ \hphantom{4 \circledcirc 0} = 12$.